Fall 2015 Midterm Q1A

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Arianna Brooks 4A
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Joined: Wed Sep 21, 2016 2:57 pm

Fall 2015 Midterm Q1A

Postby Arianna Brooks 4A » Tue Nov 01, 2016 11:23 pm

Q1A. reads:

2A + 1B ----> 3C
If one mole of A is mixed with one mole of B, which is the limiting reactant?

The answer says that A is the limiting reactant, but I really don't understand this. Wouldn't B be used up faster because there's less of it? Can someone please explain this to me? Thanks

Vista_Farkhondeh_3O
Posts: 26
Joined: Tue Jul 12, 2016 3:00 am

Re: Fall 2015 Midterm Q1A

Postby Vista_Farkhondeh_3O » Wed Nov 02, 2016 12:29 am

Because you need 2 moles of A for every one mole of B. You only have one mole of A, so technically, you'll only be using 1/2 mole of B. A runs out before B does, making A your limiting reactant.

MayaKhalil_1L
Posts: 34
Joined: Wed Sep 21, 2016 2:59 pm

Re: Fall 2015 Midterm Q1A

Postby MayaKhalil_1L » Wed Nov 02, 2016 12:42 am

You need to calculate the number of moles of A that is used based off of the moles of A and vice versa.
It's difficult to picture without numbers but essentially all you do is calculate the moles in terms of the variables.

So the questions says 1 mole of each variable is used:
1mol A x 1 mol B/2 mol A= 1/2 mol B

1 mol B x 2 mol A/ 1 mol B= 2 mol A

A is the limiting reactant because it only uses up 1/2 mol of B while B would require 2 mol of A. In other words, in order for B to be completely used it needs 2 moles of A, but A only uses up 1/2 mol of B. However, there is only 1 mol of each, so there is not enough for B to be completely used, but if A is used, there is excess B. Therefore, the reaction is dependent on A making it the limiting reactant.

You can also calculate the amount of moles of C produced for each one and see that less moles of C are produced with A, making it the limiting reactant.

1 mol A x 3 mol C/2 mol A = 1.5 mol C

1 mol B x 3 mol C/1 mol B= 3 mol C


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